Q 7.22    As shown in Fig.7.40, the two sides of a step ladder BA and CA  are 1.6mlong and
               hinged at A. A rope DE0.5m is tied half way up. A weight 40 kg  is suspended from
               a point F1.2m from B  along the ladder BA . Assuming the floor to be frictionless
               and neglecting the weight of the ladder, find the tension in the rope and forces
               exerted by the floor on the ladder. (Take g=9.8m/s^{2}).
              (Hint: Consider the equilibrium of each side of the ladder separately.)

 

                                           

 

Answers (1)

The FBD of the figure is shown below :

                                                       Rotational motion,   20149

Consider triangle ADH,

                                                  AH\ =\ \sqrt{\left ( AD^2\ -\ DH^2 \right )}                                            (AD and DH can be found using geometrical analysis.)

                                                             =\ \sqrt{\left ( 0.8^2\ -\ 0.25^2 \right )}\ =\ 0.76\ m                                  

Now we will use the equilibrium conditions :

     (i) For translational equilibrium :

                                                               N_c\ +\ N_b\ =\ mg\ =\ 40\times 9.8\ =\ 392\ N                    ......................................(i)

    (ii)  For rotational equilibrium :

                                                              -N_b\times BI\ +\ mg\times FG\ +\ N_c \times CI \ +\ T\times AG\ -\ T\times AG\ =\ 0

or                                                            (N_c\ -\ N_b)\times 0.5\ =\ 49

or                                                             N_c\ -\ N_b\ =\ 98                       ............................................................(ii)

Using (i) and (ii) we get :

                                                   N_c\ =\ 245\ N      and         N_b\ =\ 147\ N

Now calculate moment about point A :

                                               -N_b \times BI \ +\ mg\times FG\ +\ T\times AG\ =\ 0

Solve the equation :                                   T\ =\ 96.7\ N

                    

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